Question: $h(t) = t$ $g(t) = -6t^{2}-6t-5-3(h(t))$ $ h(g(-6)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(-6)$ . Then we'll know what to plug into the outer function. $g(-6) = -6(-6)^{2}+(-6)(-6)-5-3(h(-6))$ To solve for the value of $g$ , we need to solve for the value of $h(-6)$ $h(-6) = -6$ $h(-6) = -6$ That means $g(-6) = -6(-6)^{2}+(-6)(-6)-5+(-3)(-6)$ $g(-6) = -167$ Now we know that $g(-6) = -167$ . Let's solve for $h(g(-6))$ , which is $h(-167)$ $h(-167) = -167$